NEET Organic Chemistry: Some Basic Principles and Techniques Previous Year Questions with Solutions

In the 1-phenylethyl carbocation, the positively charged carbon is sp² hybridized. It forms three σ bonds in a trigonal planar geometry, leaving one unhybridized 2p orbital that is completely empty (contains 0 electrons).

• • Hybridization of C⁺: sp².
• • Acceptor orbital: vacant unhybridized 2p orbital.

The empty 2p orbital on the benzylic carbon aligns parallel to the adjacent p-orbitals of the aromatic ring. This enables delocalization of the filled aromatic π-bonding electrons into the empty p-orbital, dispersing the positive charge across the ortho and para positions of the benzene ring.

• • Donor: Filled π-bonding molecular orbitals of the benzene ring.
• • Effect: Extensive benzylic resonance stabilization.

The empty 2p orbital also interacts with the adjacent sp³-hybridized C-H σ-bonding orbitals of the methyl group. This no-bond resonance (hyperconjugation) involves the overlap of the filled C-H σ orbital with the empty p orbital.

• • Donor: Filled σ orbitals of the adjacent C-H bonds.
• • Overall Interaction: Empty p orbital interacts with filled σ and filled π orbitals.

1. Conjugated Dienes: Double bonds are separated by exactly one single bond (-C=C-C=C-). Continuous unhybridized p-orbitals allow π-electron delocalization. 2. Isolated Dienes: Double bonds are separated by two or more sp³ hybridized carbon atoms / single bonds (-C=C-(CH₂)ₙ-C=C-). 3. Cumulated Dienes (Allenes): Double bonds are adjacent to the same carbon atom (-C=C=C-).

• • Delocalization stabilizes the system.

Writing the IUPAC structure of hepta-1,3-diene: CH₂=CH-CH=CH-CH₂-CH₂-CH₃

• • The first double bond is between C1 and C2.
• • The second double bond is between C3 and C4.
• • They are separated by a single σ bond between C2 and C3.
• • Thus, hepta-1,3-diene contains conjugated double bonds.

1. hepta-1,4-diene: CH₂=CH-CH₂-CH=CH-CH₂-CH₃ (separated by one -CH₂- group; isolated). 2. hepta-1,5-diene: CH₂=CH-CH₂-CH₂-CH=CH-CH₃ (separated by two -CH₂- groups; isolated). 3. hepta-1,6-diene: CH₂=CH-CH₂-CH₂-CH₂-CH=CH₂ (separated by three -CH₂- groups; isolated).

• • Matches Option (1).

Molecular polarity depends on the dipole moment, the presence of electronegative heteroatoms (O), and above all, the ability to engage in hydrogen bonding (both as a donor and acceptor), which dictates chromatographic adsorption and dielectric properties.

• • Hydrogen bond donors (protic) are significantly more polar than aprotic species of comparable molecular weight.

1. Diethyl ether (CH₃CH₂OCH₂CH₃, A): Symmetrical ether with bent C-O-C geometry. Weak dipole moment (μ ≈ 1.15 D), lacks hydrogen-bonding protons. Least polar of the four. 2. Acetone (CH₃COCH₃, C): Polar aprotic ketone containing a strongly polarized carbonyl group (>C=O, μ ≈ 2.88 D). More polar than ether, but cannot donate hydrogen bonds. 3. Ethanol (CH₃CH₂OH, B): Polar protic alcohol. Contains a highly polarized O-H bond capable of acting as both a strong hydrogen bond donor and acceptor. 4. Acetic acid (CH₃COOH, D): Contains both a carbonyl (>C=O) and a highly polarized hydroxyl (-O-H) group. Readily forms cyclic hydrogen-bonded dimers and exhibits strong ion-dipole/polar interactions, making it the most polar.

• • Increasing order: A < C < B < D.

The correct sequence in increasing order of polarity is A < C < B < D.

• • Matches Option (4).

Examine separation methods based on boiling point differentials:

• • A matches IV (A-IV).
• • Fractional Distillation (III): Used when boiling point differences are small (fractions of crude oil in petroleum refining).
• • B matches III (B-III).

Examine specialized distillation techniques for delicate or steam-volatile compounds:

• • Glycerol boils at 563 K with decomposition, but boils at 453 K without decomposition under 12 mm Hg pressure; it is separated from spent-lye by vacuum distillation (C-I).
• • Steam Distillation (II): Used for substances which are steam volatile and immiscible with water (such as the aniline-water mixture, D-II).

Assemble all pairs:

• • Matches Option (1) with 100% textbook precision.

Nitrogen, sulfur, halogens, and phosphorus present in an organic compound are covalently bonded. To detect them, they must be converted into ionic inorganic salts by fusing the compound with metallic sodium in an ignition tube:

• • Reactions (1), (2), and (3) are the three foundational fusion reactions of Lassaigne’s test.

The detection of carbon and hydrogen is performed separately by heating the organic compound with dry cupric oxide (CuO) in a hard glass test tube:

• • The evolved CO₂ is detected by passing it through lime water (which turns milky), and H₂O is detected by anhydrous copper sulfate (turning blue).
• • This test does not use metallic sodium and is completely distinct from Lassaigne’s test.

Reaction (4) belongs to the detection of carbon via copper oxide and does NOT belong to Lassaigne’s test:

• • Matches Option (4) perfectly.

Sublimation is a phase transition that occurs at temperatures and pressures below a substance’s triple point in its phase diagram, where the solid phase’s vapour pressure equals atmospheric pressure before melting occurs. When heated, the solid transitions directly into vapour without entering the liquid state:

Sublimation is employed to separate sublimable solids (e.g., naphthalene, anthracene, benzoic acid, camphor, ammonium chloride, iodine) from non-sublimable impurities. On cooling, the pure vapours condense directly into solid form on a chilled surface (such as an inverted funnel plugged with cotton), leaving non-volatile impurities behind in the dish.

A carbocation is an sp²-hybridized, planar, sextet electron-deficient carbon. Stability is enhanced by electron donation through: 1. Inductive effect (+I): Alkyl groups push electron density toward the positive center. 2. Hyperconjugation: Delocalization of σ(C-H) electrons of α-carbons into the empty 2pz orbital of the carbocation. The more α-hydrogens, the greater the number of hyperconjugative resonance structures.

Evaluating the α-hydrogen count and degree of each carbocation: - Option (1): Cyclopentyl-CH2⁺: Primary (1°) carbocation. Only 1 α-hydrogen (at the methine bridgehead). Very unstable. - Option (2): 1-Methylcyclohexyl cation: Tertiary (3°) carbocation. C-1 is bonded to three carbon atoms: the exocyclic methyl carbon (3 α-H) and two endocyclic methylene carbons at C-2 (2 α-H) and C-6 (2 α-H). Total α-H = 3 + 2 + 2 = 7 α-hydrogens! - Option (3): (CH3)2CH-C⁺H-CH2CH3: Secondary (2°) carbocation. α-hydrogens = 1 (from isopropyl CH) + 2 (from ethyl CH2) = 3 α-hydrogens. - Option (4): CH3-C⁺H-CH2-CH(CH3)2: Secondary (2°) carbocation. α-hydrogens = 3 (from terminal methyl) + 2 (from methylene CH2) = 5 α-hydrogens.

Structure (2) is the only tertiary (3°) carbocation in the set. Its tertiary status combined with 7 hyperconjugative α-hydrogens makes it by far the most stable species.

During the preparation of Lassaigne’s extract, the organic compound is fused with molten metallic sodium. When both nitrogen and sulphur are present in the organic framework, they react jointly with carbon and sodium to form sodium thiocyanate rather than separate sodium cyanide and sodium sulphide:

The fused mass is boiled with distilled water and filtered. The resulting alkaline sodium fusion extract contains soluble sodium thiocyanate (NaSCN), dissociating into sodium cations and thiocyanate anions:

When acidified with dilute hydrochloric acid and treated with a few drops of ferric chloride solution, ferric ions (Fe³⁺) coordinate with thiocyanate ions to form the soluble blood-red colored coordination complex ion:

Prussian blue, Fe₄[Fe(CN)₆]₃ · xH₂O, requires free cyanide ions (CN⁻). Because nitrogen is tied up as SCN⁻, no Prussian blue forms unless an excess of sodium is used to break thiocyanate into cyanide and sulphide.

Pyridine has the chemical formula C₅H₅N. It consists of a planar hexagonal ring of 5 carbon atoms and 1 nitrogen atom, with each carbon bonded to one hydrogen atom:

Sum the ring skeletal bonds and the peripheral C–H single bonds:

The aromatic sextet consists of 6 delocalized π electrons distributed across 3 conjugated double bonds:

The nitrogen atom in pyridine is sp² hybridized. One unhybridized p_z orbital contributes 1 electron to the aromatic π cloud, while the third sp² hybrid orbital holds an unshared lone pair in the ring plane:

According to NCERT, metamerism is a type of constitutional isomerism that arises due to different alkyl chains attached to the same polyvalent functional group (e.g., ether $-O-$, thioether $-S-$, secondary amine $-NH-$, or ketone $-CO-$) in the molecule.

The molecular formula $C_4H_{10}O$ corresponds to saturated mono-oxygenated compounds (degree of unsaturation $DBE = 0$). Among the isomeric ethers, we can distribute the 4 carbons on either side of the divalent oxygen atom in multiple distinct ways: 1. Ethoxyethane (Diethyl ether): $CH_3-CH_2-O-CH_2-CH_3$ (symmetrical $2 + 2$ split) 2. 1-Methoxypropane (Methyl propyl ether): $CH_3-O-CH_2-CH_2-CH_3$ ($1 + 3$ linear split) 3. 2-Methoxypropane (Methyl isopropyl ether): $CH_3-O-CH(CH_3)_2$ ($1 + 3$ branched split) These compounds belong to the same homologous family and have identical functional groups, but differ in the alkyl chains flanking the oxygen atom. They are metamers.

- $C_5H_{12}$: Saturated pentane; exhibits only chain isomerism (pentane, isopentane, neopentane) as it contains no polyvalent heteroatom. - $C_3H_8O$: Ethers require dividing 3 carbons across oxygen; the only possible division is $1 + 2$ ($CH_3-O-C_2H_5$). No second ether isomer exists. - $C_3H_6O$: Ketone is propan-2-one ($CH_3-CO-CH_3$), which cannot have unequal alkyl groups since only 2 carbons remain to flank the carbonyl carbon.

Based on the principle of separation, chromatography is classified into two broad categories: 1. Adsorption chromatography 2. Partition chromatography.

Adsorption chromatography is based on the differential adsorption of components on a solid adsorbent (silica gel or alumina). It encompasses: (a) Column chromatography (b) Thin layer chromatography (TLC).

Partition chromatography is based on continuous differential partitioning of components between stationary and mobile phases. In Paper Chromatography, a special chromatography paper contains water trapped in the cellulose fibers which acts as the stationary liquid phase. A strip of chromatography paper spotted at the base is suspended in a suitable mobile solvent which rises by capillary action.

NCERT explicitly highlights: "Paper chromatography is an example of partition chromatography."

Carbocations possess a sp2-hybridized planar central carbon with an unhybridized, empty p-orbital perpendicular to the molecular plane.

Hyperconjugation involves the overlap of $$\sigma$$-electrons of adjacent $$\text{C}-\text{H}$$ bonds of the alkyl group with the empty $$p$$-orbital of the carbocationic carbon (also called no-bond resonance). This delocalizes the positive charge across multiple hydrogen atoms, dispersing charge and stabilizing the cation.

- Tertiary butyl carbocation, $$(\text{CH}_3)_3\text{C}^+$$, has three methyl groups directly attached to the carbocationic carbon, giving a total of $$3 \times 3 = 9\text{ }\alpha\text{-hydrogens}$$, yielding 9 hyperconjugative canonical structures. - Secondary butyl carbocation, $$\text{CH}_3-\text{CH}^+-\text{CH}_2-\text{CH}_3$$, has one methyl ($$3\text{ }\alpha\text{-H}$$) and one methylene ($$2\text{ }\alpha\text{-H}$$), totaling only $$5\text{ }\alpha\text{-hydrogens}$$, yielding 5 hyperconjugative structures.

- Methyl groups exhibit $$+I$$ effect, NOT $$-I$$ effect (Option 1 is false). - Methyl groups lack lone pairs or $$\pi$$-bonds and CANNOT exhibit resonance ($$+R$$ or $$-R$$) effects (Options 2 and 3 are chemically invalid). - Thus, Hyperconjugation uniquely and correctly accounts for the superior stability of tert-butyl carbocation.